Two spherical bodies of mass M and 5M and radii R and 2R are released in free space with initial separation between their centres equal to 12R. If they attract each other due to gravitational force only, then the distance covered by the smaller body before collision is:

Assuming velocities of masses M and 5M are v1 and v2 respectively
Using conservation of momentum principle
Mv1 + 5Mv2 = 0 → v2 = -v1/5.. (i)
Using energy conservation principle
ΔP.E. = ΔK.E.
The P.E. is given as Gm1m2/r
-G5M²/12R - (-G5M²/3R) = 1/2Mv1² + 1/2(5M)v2²
From equation (i) substituting value of v2
5GM²/12R = 1/2Mv1² + 1/2(5M)(-v1/5)²
5GM²/4R = 1/2Mv1² + 1/10Mv1²
12Mv1² = 5GM²/4R
1/2Mv1² = K.E. of mass M just before collision = 25GM²/24R
Using work energy theorem: work done by gravitational force = ΔK.E.
∫₁₂RF.dr = 1/2mv1²
∫₁₂R GM5M/r²dr = 1/2mv1²
[GM5M/r]₁₂R = 1/2mv1²
GM5M(1/r - 1/12R) = 25GM²/24R → r = 7.5R
hence correct answer is option A